Cross Product of Elements of Standard Ordered Basis
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Theorem
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
- $\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$
and:
\(\ds \mathbf i \times \mathbf j\) | \(=\) | \(\, \ds \mathbf k \, \) | \(\, \ds = \, \) | \(\ds -\mathbf j \times \mathbf i\) | ||||||||||
\(\ds \mathbf j \times \mathbf k\) | \(=\) | \(\, \ds \mathbf i \, \) | \(\, \ds = \, \) | \(\ds -\mathbf k \times \mathbf j\) | ||||||||||
\(\ds \mathbf k \times \mathbf i\) | \(=\) | \(\, \ds \mathbf j \, \) | \(\, \ds = \, \) | \(\ds -\mathbf i \times \mathbf k\) |
where $\times$ denotes the dot product.
Proof
From Cross Product of Vector with Itself is Zero:
- $\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$
Then we can take the definition of cross product:
- $\mathbf a \times \mathbf b = \begin {vmatrix}
\mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \\ \end {vmatrix} = \mathbf a \times \mathbf b = \paren {a_j b_k - a_k b_j} \mathbf i - \paren {a_i b_k - a_k b_i} \mathbf j + \paren {a_i b_j - a_j b_i} \mathbf k$
and note that:
\(\ds \mathbf i\) | \(=\) | \(\ds 1 \mathbf i + 0 \mathbf j + 0 \mathbf k\) | ||||||||||||
\(\ds \mathbf j\) | \(=\) | \(\ds 0 \mathbf i + 1 \mathbf j + 0 \mathbf k\) | ||||||||||||
\(\ds \mathbf k\) | \(=\) | \(\ds 0 \mathbf i + 0 \mathbf j + 1 \mathbf k\) |
Hence:
\(\ds \mathbf i \times \mathbf j\) | \(=\) | \(\ds \paren {0 \cdot 0 - 0 \cdot 1} \mathbf i - \paren {1 \cdot 0 - 0 \cdot 0} \mathbf j + \paren {1 \cdot 1 - 0 \cdot 0} \mathbf k\) | \(\ds = \mathbf k\) | |||||||||||
\(\ds \mathbf j \times \mathbf k\) | \(=\) | \(\ds \paren {1 \cdot 1 - 0 \cdot 0} \mathbf i - \paren {0 \cdot 1 - 0 \cdot 0} \mathbf j + \paren {0 \cdot 0 - 1 \cdot 0} \mathbf k\) | \(\ds = \mathbf i\) | |||||||||||
\(\ds \mathbf k \times \mathbf i\) | \(=\) | \(\ds \paren {0 \cdot 0 - 1 \cdot 0} \mathbf i - \paren {0 \cdot 0 - 1 \cdot 1} \mathbf j + \paren {0 \cdot 0 - 0 \cdot 1} \mathbf k\) | \(\ds = \mathbf j\) |
The remaining identities follow from Vector Cross Product is Anticommutative:
\(\ds \mathbf i \times \mathbf j\) | \(=\) | \(\ds -\mathbf j \times \mathbf i\) | ||||||||||||
\(\ds \mathbf j \times \mathbf k\) | \(=\) | \(\ds -\mathbf k \times \mathbf j\) | ||||||||||||
\(\ds \mathbf k \times \mathbf i\) | \(=\) | \(\ds -\mathbf i \times \mathbf k\) |
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $4$. The Vector Product: $(2.16)$, $(2.17)$
- 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 3$: $(7)$, $(8)$